n 8:=

0.Initial data:

Choose C25/30

S500

XC1/S4

fck 25N

mm2

:=

fcd 16.67N

mm2

:=

fyd 435N

mm2

:=

Ecm 31GPa:=

a 6.8m 0.05m n⋅+ 7.2m=:=b a 1.2m− 0.025m n− 5.8 m=:=

Location :

ks 0.2g:=

1. Values of loadings

Loadings at column's base:

Fundamental series(GF):•Med1 180kN m⋅ 5kN m⋅ n⋅+ 220 kN m⋅⋅=:=

Med2 185kN m⋅ 10kN m⋅ n⋅+ 265 kN m⋅⋅=:=

Medf Med2 265 kN m⋅⋅=:=

Ned1 1670kN 5kN n⋅+ 1.71 103

× kN⋅=:=

Ned2 1900kN 10kN+ 1.91 103

× kN⋅=:=

Nedf Ned2 1.91 103

× kN⋅=:=

Seismic series(GS):•Med3 185kN m⋅ 10kN m⋅ n⋅+ 265 kN m⋅⋅=:=

Med4 200kN m⋅ 5kN m⋅ n⋅+ 240 kN m⋅⋅=:=

Meds Med3 265 kN m⋅⋅=:=

Ned3 1300kN 10kN n⋅+ 1.38 103

× kN⋅=:=

Ned4 1280kN 5kN n⋅+ 1.32 103

× kN⋅=:=

Neds Ned3 1.38 103

× kN⋅=:=

Vedmax 90kN 5kN n⋅+ 130 kN⋅=:=

Loadings of the cantilever(at 20cm from the column's side)

V1 180kN 2.5kN n⋅+ 200 kN⋅=:=H1 0.1 V1⋅ 20 kN⋅=:=

2.Dimensions

Column's length

l0.6− m 1.5m−( )−

2a+ 8.25 m=:=

Buckling length:

l0 2l 16.5 m=:=

Predesign of the column:

-technological condition: multiple of 5cm

-fire resistance condition:

for R120(120 min exposure until colapse),multiple sides exposure and µg=0.2 (tab.6.2-Kiss):

bc 20cm:=hc 20cm:=and the concrete cover a=4cm

-minimum dimension condition:

bc 30cm≥hc 30cm≥

-resistance condition:

for end column:

n1 1.6:=

n1Nedf

0.8 fcd⋅⋅ 2.292 10

3× cm

2⋅=

So min bc x hc = 2292 cm2

Choose

bc 50cm:=hc 50cm:=

bc hc⋅ 2.5 103

× cm2

⋅=

-normalised axial force condition:

for :

bc 50cm:=hc 50cm:=

vd

Neds

fcd bc⋅ hc⋅0.331=:=

vd 0.4≤

-flexibility condition:

λlim 70:=

So,we have an intermediary column with 50x50 cross-section.

3.Moment increase due additional eccentricity

Eccentricity ea 20mm:=

MEd.1 Med1 Ned1 ea⋅+ 254.2 kN m⋅⋅=:=

MEd.2 Med2 Ned2 ea⋅+ 303.2 kN m⋅⋅=:=

MEd.3 Med3 Ned3 ea⋅+ 292.6 kN m⋅⋅=:=

MEd.4 Med4 Ned4 ea⋅+ 266.4 kN m⋅⋅=:=

4.Moment increase due second order effects

Approximative reinforcement

Area of cross-section:

Acs bc hc⋅ 2.5 103

× cm2

⋅=:=

Approximative reinforcemet area:

p1.5

100Acs⋅ 37.5 cm

2⋅=:=

Choose 8Φ25 with area

As 39.28cm2

:=

number of bars

n 8:=diameter :

ϕ 25mm:=

concrete cover

XC1:

cmindur 15mm:=

cnom cmindur 10mm+ 25 mm⋅=:=

d1 cnomϕ

2+ 37.5 mm⋅=:=

take

d1 40mm:=

Nominal rigidity

Isn π ϕ

2⋅( )⋅

4d1

hc

2−

⋅ d1hc

2−

1.732 104−

× m4

⋅=:=

Ic bchc

3

12⋅ 5.208 10

3−× m

4=:=

ρAs

Acs0.016=:=

ρ 0.01≥

Ks 0:=

Eurocode 1992-1-1 3.1.4

h0 2Acs

2 bc hc+( )⋅ 0.25 m=:=

ϕef 2.8:=

Kc 0.3 1 0.5ϕef+( ) 0.72=:=

EcdEcm

1.225.833 GPa⋅=:=

Es 200GPa:= 3.2.7 (4)

Nominal rigidity:

EI Kc Ecd⋅ Ic⋅ Ks Es⋅ Is⋅+ 9.688 104

× kN m2

⋅⋅=:=

Moments amplification coefficient

c0 8:=

βπ( )

2

c0

1.234=:=

NB π2 EI

l02

⋅ 3.512 103

× kN⋅=:=

Med.1 MEd.1 1β

NB

Ned1

1−

+

⋅ 551.81 kN m⋅⋅=:=

Med.2 MEd.2 1β

NB

Ned2

1−

+

⋅ 749.199 kN m⋅⋅=:=

Med.3 MEd.3 1β

NB

Ned3

1−

+

⋅ 526.265 kN m⋅⋅=:=

Med.4 MEd.4 1β

NB

Ned4

1−

+

⋅ 464.322 kN m⋅⋅=:=

5.Dimensioning of the longitudinal reinf.

µMed.2

bc hc2

⋅ fcd⋅

0.36=:=

υNedf

bc hc⋅ fcd⋅0.458=:=

d1 d2=

d1

hc0.08=

Fig. 3.6 b) (Kiss)

ωtot 0.60:=

Astot ωtot bc⋅ hc⋅fcd

fyd

⋅ 57.483 cm2

⋅=:=

Because from the approximative design it resulted a smaller value,we will use these datas.

As 64.32cm2

:=

number of bars

n 8=diameter :

ϕ 32m:=

d1 40 mm⋅=

Verifiying the minimum distance between reinforcements:

dr

hc 2 d1⋅−( )3

140 mm⋅=:=

dr ϕ>

6.Dimensioning of transversal reinforcement

Ved Vedmax 130 kN⋅=:=

Capable shear force•

VRdc = CRd.c η1⋅ k 100ρl fck⋅( )1

3⋅

⋅ k1 σcp⋅+

bc⋅ d⋅

d hc d1− 0.46 m=:=

γc 1.5:=CRd.c

0.18

γc

0.12=:=

η1 1:= -for heavy concrete

k 1200mm

d+ 1.659=:= < 2.00

ρlAs

bc d⋅0.02797=:= < 0.02

σcp

Nedf

bc hc⋅7.64

N

mm2

⋅=:=

k1 0.15:=

fck

fck

N

mm2

25=:= σcp

σcp

N

mm2

7.64=:=

dd

mm460=:=

bcbc

mm500=:=

VRdc CRd.c η1⋅ k 100ρl fck⋅( )1

3⋅

⋅ k1 σcp⋅+

bc⋅ d⋅ 1⋅ N 452.252 kN⋅=:=

VRdc 452.252 kN⋅=

Minimum capable shear force:

VRdc.min = νmin k1 σcp⋅+( ) bc⋅ d⋅

νmin 0.035 k

3

2⋅ fck⋅ 0.374=:=

VRdc.min νmin k1 σcp⋅+( ) bc⋅ d⋅ N⋅ 349.62 kN⋅=:=

Ved 130 kN⋅= < VRdc.min 349.617 kN⋅= < VRdc 452.252 kN⋅=

trasversal reinforcement will be constructive

Choose transversal reinforcement diameter:

ϕw 10mm:=

2 stirrups=>4 shear branches

Asw 4π ϕw

2⋅

4

⋅ 314.16 mm2

⋅=:=

Distance between stirrups in the critical zones

ϕ 3.2 104

× mm⋅=

bc 500mm:=

scr.max min 6 ϕ⋅bc 2 d1⋅−

3, 125mm,

125 mm⋅=:=

take

scr.max 120mm:=

ρw.eff

Asw

bc scr.max⋅0.00524=:=

ρw.min 0.005:=

ρw.eff ρw.min≥

it's OK , the minimum percentage of reinforcement is exceeded

Length of the critical zone:

l 8.25 m=

lcr max 1.5bcl

6, 600mm,

1.375 m⋅=:=

take

lcr 1.56m:=

Distance between stirrups in the non-critical zones

sncr.max min 20 ϕ⋅ min bc hc, ( ), 400mm, ( ) 400 mm⋅=:=

take

sncr.max 300mm:=

So:

-on the critical lengths=1.56m from the ground level Φ10/12

-under ground level,in the foundation zone, same Φ10/12

-in rest Φ10/30

7.Design of the short cantilever

distance from application point of the force to the connection point of the column

ac 20cm:=

Fed V1 200 kN⋅=:=

Hed H1 20 kN⋅=:=

bcant - cantilever thickness,taken as long as column thickness

Take :

bcant bc 500 mm⋅=:=

hcant 150mm:=

dcant 0.9hcant 135 mm⋅=:=

fcd 16.67N

mm2

⋅=

Longitudinal reinforcement:

M Fed ac⋅ Hed hcant⋅+ 43 kN m⋅⋅=:=

µM

bcant dcant2

⋅ fcd⋅

0.283=:=

ω 0.3412:=

Longitudinal reinforcement area will be maximum from the following two:

Asl ω bcant⋅ hcant⋅fcd

fyd

⋅ 9.807 cm2

⋅=:=

Approximative :

Asa0.15

100bcant⋅ hcant⋅ 1.125 cm

2⋅=:=

so

Asl 9.807 cm2

⋅=

take 4Φ18

Asl 10.16cm2

:=

Stirrups :

ac

hcant

1.333=

ac

hcant

0.5>

we will have horizontal stirrups and vertical stirrups

horizontal stirrups:•

Ash 0.25 Asl⋅ 2.54 cm2

⋅=:=

take 4Φ10

Ash 3.14cm2

:=

vertical stirrups:•

Asv 0.5 Asl⋅ 5.08 cm2

⋅=:=

take 5Φ12

Asv 5.65cm2

:=

distance between stirrups:•take

ds cmindur10

2mm+ 20 mm⋅=:=

ds 20mm:=

-distance between horizontal stirrups:

sh

hcant 2 ds⋅−( )3

36.667 mm⋅=:=

approx. 40mm

-distance between vertical stirrups:

take the length of the cantilever as:

lcant 300mm:=

( )

sv

lcant ds−( )4

70 mm⋅=:=

8.Column verification for manipulation

distance from both ends to the point where the column is rooted

x 0.2 l⋅ 1.65 m=:=

dead load:

gcol Acs 25⋅kN

m3

6.251

mkN⋅=:=

V1 gcoll

2⋅ 25.781 kN⋅=:=

Mr 8.62kN m⋅:=Mc 10.77kN m⋅:=

d 460mm:=z 0.9d 414 mm⋅=:=

µmMc

bc d2

⋅ fcd⋅

6.107 103−

×=:=

ωm 0.01:=

Asl.h ωm bc⋅ d⋅fcd

fyd

⋅ 0.881 cm2

⋅=:= Satisfied

Ved 130 kN⋅=

VRdc.min 349.617 kN⋅=

VRdc 452.252 kN⋅=

ctgθ 1VRdc.min Ved−( )

VRdc Ved−1.5⋅+ 2.022=:=

s1.h

Asw z⋅ fyd⋅ ctgθ⋅( )V1

4.438 m⋅=:= Satisfied

9. Checking column at assembling

(l-x)V1+gcol*x*x/2-gcol*(l-x)*(l-x)/2=0

l 8.25 m=

x 1.65 m=

gcol 6.251

mkN⋅=

V1 19.59kN:=

Mc 30.30kN m⋅:=

Mr 8.62kN m⋅:=

µmMc

bc d2

⋅ fcd⋅

0.017=:=

ωm 0.0171:=

Asl.h ωm bc⋅ d⋅fcd

fyd

⋅ 1.507 cm2

⋅=:= Satisfied

s1.h

Asw z⋅ fyd⋅ ctgθ⋅( )V1

5.84 m⋅=:= Satisfied

Mark Diameter Type Length(m) Quantity 8 10 18 32

1 32 S500 8,62 8 68,96

2 18 S500 1,50 4 6,00

3 10 S500 1,68 47 78,96

4 10 S500 1,17 47 54,99

5 10 S500 2,42 4 9,68

6 12 S500 1,14 5 5,7

9,68 139,65 6,00 68,96

0,39 0,62 2,00 6,31

3,82 86,06 11,98 435,15

Total 537,00

Total length(m)

Reinforcement extras

Length by diameter

Total weight(kg)

Weight for diameter(kg/m)